Functional Equation

Find all functions $f: \mathbb R \to \mathbb R$ such that for arbitrary real numbers $x$ and $y$ ,

$f(x+y) + f(x) f(y) = f(xy) + f(x) + f(y) .$

#Algebra

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Comments

Put $x=y=0$ $f(0)+f(0)^{2}=3f(0)$ This implies $f(0)=0$ or $f(0)=2$ .

CaseI : $f(0)=2$ Now put $x=0$ $3f(y)=4+f(y)$ $\boxed{f(y)=2}$

Case II: $f(0)=0$ Put $x=y=2$ $f(4)+f(2)^{2}=f(4)+2f(2)$ $f(2)=2$ or $f(2)=0$ .

Since the main functional equation all terms are only monic functions so another solution is $f(y)=0$ .

Other solution is $f(y)=y$ which can be proved by induction.

Shivam Jadhav - 5 years, 5 months ago

You haven't shown that there is no other solution. In fact, as my solution suggests, if we restrict $f$ only to the integers, there exists another solution, and you haven't ruled it out.

Ivan Koswara - 5 years, 5 months ago

Yea Ivan has a point. I don't understand how u can use induction when the domain is reals ?

Shrihari B - 5 years, 5 months ago

The induction part is wrong, but it's a minor mistake; verifying that $f(x) = x$ works is simply plugging it into the equation. The problem is that other potential solutions haven't been ruled out.

Ivan Koswara - 5 years, 5 months ago

As mentioned by Ivan, I don't understand what you are doing here. I agree up the Case 2, but fail to follow form "Since the main functional equation ..."

Calvin Lin Staff - 5 years, 5 months ago

I mean that both sides have terms which have 1 as coefficient so each term can be zero.

Shivam Jadhav - 5 years, 5 months ago

Let $P(x,y)$ be the statement $f(x+y) + f(x)f(y) = f(xy) + f(x) + f(y)$ .

$P(x,0) \implies f(x) + f(x)f(0) = f(0) + f(x) + f(0) \implies f(x)f(0) = 2f(0)$ . If $f(0) \neq 0$ , we can cancel it to give $f(x) = 2$ for all $x$ , which is a solution. Otherwise, $f(0) = 0$ .

$P(2,2) \implies f(4) + f(2)f(2) = f(4) + f(2) + f(2) \implies f(2)^2 = 2f(2)$ . This gives $f(2) = 0$ or $f(2) = 2$ .

$P(1,1) \implies f(2) + f(1)^2 = 3f(1)$ . If $f(2) = 0$ , this leads to $f(1) = 0$ or $f(1) = 3$ . If $f(2) = 2$ , this leads to $f(1) = 1$ or $f(1) = 2$ .

$P(-1,1) \implies f(-1)f(1) = 2f(-1) + f(1)$ . $f(1) = 2$ leads to a contradiction, so we are left with three cases. To recap, our current cases are $(f(1), f(2)) = (0,0), (1,2), (3,0)$ .

$P(x,1) \implies f(x+1) + f(x)f(1) = 2f(x) + f(1)$ . We can now apply induction (in both ways, going on the positive integers and the negative integers) to determine the value of $f$ on the integers:

Case 1: $f(1) = 0,$ . Then $P(x,1) \implies f(x+1) = 2f(x)$ . Inducting gives $f(x) = 0$ for all integer $x$ .

Case 2: $f(1) = 1$ . Then $P(x,1) \implies f(x+1) = f(x) + 1$ . Inducting gives $f(x) = x$ for all integer $x$ .

Case 3: $f(1) = 3$ . Then $P(x,1) \implies f(x+1) = 3 - f(x)$ . Inducting gives $f(x) = 3$ if $x$ is odd and $f(x) = 0$ if $x$ is even.

Let $x = \frac{p+1}{p}, y = p+1$ for nonzero integer $p$ . Note that $x+y = xy$ . Thus $P(x,y) \implies f(x)f(y) = f(x) + f(y)$ . Since we know the value of $f(y)$ , we can determine the value of $f(x)$ . That is, the value of $f \left( 1 + \frac{p}{1} \right)$ .

Using $P(x,1)$ , we can induct again to determine the values of $f \left( n + \frac{1}{p} \right)$ for all integer $n$ , including $n = 0$ .

$P \left( \frac{1}{p}, n \right) \implies f \left( \frac{n}{p} \right) + f \left( \frac{1}{p} \right) f(n) = f \left( n + \frac{1}{p} \right) + f \left( \frac{1}{p} \right) + f(n)$ . Since we know the values of $f \left( \frac{1}{p} \right)$ , $f \left( n + \frac{1}{p} \right)$ , and $f(n)$ , we know the value of $f \left( \frac{n}{p} \right)$ . That is, we know $f$ to the rationals. I haven't actually worked it out to figure out whether Case 3 still persists.

I'm not sure how to extend this to the reals. I'm also pretty sure this is too convoluted (that is, there should be a simpler solution). You might want to first try to solve the following: find all $f$ on the reals such that $f(x+y) = f(x) + f(y)$ and $f(xy) = f(x)f(y)$ for all reals $x,y$ .

Ivan Koswara - 5 years, 5 months ago

I believe that you're missing the constant solution $f(x) = 2$ .

Calvin Lin Staff - 5 years, 5 months ago

I already wrote that one near the top.

Ivan Koswara - 5 years, 5 months ago

Can u too please post some functional equations to solve ? I am quite weak at that

Shrihari B - 5 years, 5 months ago

Try these:

$1$ . If $f$ is a polynomial function satisying $2+f(x)f(y)=f(x)+f(y)+f(xy)$ for all real $x,y$ and if $f(2)=5$ , find $f(f(2))$ .

$2$ . A polynomial function $f(x)$ satisfies the condition $f(x)f(\dfrac{1}{x})=f(x)+f(\dfrac{1}{x})$ if $f(12)=1729$ , then find $f(10)$ .

$3$ .Find all functions $f$ from $R{0,1}$ to $R$ satisfying the functional relation $f(x)+f(\dfrac{1}{1-x})=\dfrac{2(1-2x)}{x(1-x)}$ . Source:INMO

Buy the book Functional Equations by B.J. Venkatachala. It has very nice functional equations problem .

A Former Brilliant Member - 5 years, 5 months ago

A functional equation problem shouldn't have the statement "if $f$ is a polynomial..."; that statement makes it a polynomial equation problem (which has quite a different method to solve them).

Ivan Koswara - 5 years, 5 months ago

Hey should we make a thread for "Brilliant Functional Equation Contest" ? This is just an excuse for me to improve our functional equation as one problem in INMO is definitely on functional equations.

Shrihari B - 5 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$